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Equations
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1387 Accepted Submission(s): 542

Problem Description
Consider equations having the following form:

a*x1^2+b*x2^2+c*x3^2+d*x4^2=0
a, b, c, d are integers from the interval [-50,50] and any of them cannot be 0.

It is consider a solution a system ( x1,x2,x3,x4 ) that verifies the equation, xi is an integer from [-100,100] and xi != 0, any i ∈{1,2,3,4}.

Determine how many solutions satisfy the given equation.

Input
The input consists of several test cases. Each test case consists of a single line containing the 4 coefficients a, b, c, d, separated by one or more blanks.
End of file.

Output
For each test case, output a single line containing the number of the solutions.

Sample Input
1 2 3 -4
1 1 1 1

Sample Output
39088
0

#include <iostream><br /> using namespace std;<br /> int base[101];</p> <p>int hash[2000009];</p> <p>int main()<br /> {<br /> int a,b,c,d;<br /> int i,j,k;<br /> for(i=0;i<101;i++)<br /> base[i]=i*i;</p> <p> while(scanf("%d %d %d %d",&a,&b,&c,&d)!=EOF)<br /> {</p> <p> if( (a>0 && b>0 && c>0 && d>0 )|| (a<0 && b<0 && c<0 && d<0) )<br /> { printf("0/n");<br /> continue;<br /> }<br /> else<br /> {<br /> memset(hash,0,sizeof(hash));<br /> for(i=1;i<101;i++)<br /> for(j=1;j<101;j++)<br /> hash[base[i]*a+base[j]*b+1000000]++;</p> <p> int sum=0;</p> <p> for(i=1;i<101;i++)<br /> for(j=1;j<101;j++)<br /> sum+=hash[-(c*base[i]+d*base[j])+1000000];</p> <p> cout<<sum*16<<endl; //X分正负，每个X有2个解，4个就有16种不同的解<br /> }</p> <p> }<br /> }